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Afina-wow [57]
3 years ago
12

How does the electron structure of atoms change when they form chemical bonds

Chemistry
1 answer:
andriy [413]3 years ago
3 0
For ionic bond
The metal atom will lose electrons to form cations and the non metal atom will gain electron.egNaCl
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Which organic compound is correctly matched with the subunit that composes it?
Alex787 [66]
It is the Starch-glucose. Glucose is a solitary sugar particle that your body can retain specifically in the digestive system. Sucrose and starches are starches shaped by at least two sugars reinforced together. The sugars in sucrose and starch must be separated into glucose particles in the gastrointestinal tract before your digestive organs can assimilate them.
7 0
3 years ago
Can someone please answer this for me I need help I’ll give brainliest :((((
Ahat [919]
B) As water temp increases solubility increases
6 0
3 years ago
List at least two metric units that you used during the measurement activity to represent volume
Oliga [24]

Answer:

ml and cm3

Explanation:

millilitres and centimetres cubed or litres and metres cubed. Any unit of measuring liquid if the substance is a a liquid or a unit cubed

3 0
3 years ago
Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of ammonium phosphate and nickel(
densk [106]

Answer:

2PO_4^{3-}(aq)+3Ni^{2+}(aq)\rightarrow Ni_3(PO_4)_2(s)

Explanation:

Hello,

In this case, the molecular chemical reaction is:

(NH_4)_3PO_4(aq)+NiBr_2(aq)\rightarrow NH_4Br(aq)+Ni_3(PO_4)_2(s)

Thus, by considering the ionic species we have:

NH_4^+(aq)+2PO_4^{3-}(aq)+3Ni^{2+}(aq)+Br^-(aq)\rightarrow NH_4^+(aq)+Br^-(aq)+Ni_3(PO_4)_2(s)

Wherein the ammonium and bromide ions are spectator ions since nickel (II) phosphate is largely insoluble in water, therefore, the net ionic equation is:

2PO_4^{3-}(aq)+3Ni^{2+}(aq)\rightarrow Ni_3(PO_4)_2(s)

Best regards.

7 0
3 years ago
Determine the mass of ammonium chloride, NH4CI, required to prepare 0.250 L of a 0.35 M solution of ammonium chloride.
kkurt [141]
A 0.250 L solution of NH4Cl contains (0.250 L)(0.35 mol/L) = 0.0875 moles of dissolved NH4Cl. Converting this quantity to mass, we get (0.0875 mol NH4Cl)(53.491 g/mol) = 4.7 g NH4Cl (two significant figures).

So, approximately 4.7 g of NH4Cl dissolved in a 0.250 L solution will have a concentration of 0.35 L.
7 0
2 years ago
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